# Help with Laplace Transformations and 2nd order ODEs

 P: 774 For 1/(s+1)^3, let's just look at 1/s^3 ; we know that t^2 is 2/s^3 so 1/s^3 will be the inverse of (1/2)(t^2) right? but, we dont want 1/s^3, we want 1/(s+1)^3, so what do we do (like we did before)?
P: 1,031
First Shft Theorem

 Ok, the first shift theorem says that: if you have a function that looks like: e^(at)F(t); the laplace of that is the laplace of F(t) but instead of putting p, you put p-a eg: L(e^3t(t)) we know that laplace of t is 1/p^2 (you know that from the tables right?) and so here a=3 so the laplace of e^(3t)(t)=1/(p-3)^2 so, where ever you have p, you replace with p-a in this case, p-3 Do you agree??
$$t^2 = 2/s^3$$

$$\frac{1}{2}t^2 = \frac{1}{s^3}$$

using FST

$$\frac{1}{2}t^2 = \frac{1}{b^3}$$, where b = s + 1

would this make it:

\frac{1}{2}(t+1)^2 = \frac{1}{b^3} [/tex]

???

TFM
 P: 774 I dont see any e's in your answer (there should be). now as we did before, the second fraction is the laplace transform of: 4te^(-2t) because if you find the laplace of 4t, this will give you 4/s^2 but we dont want s at the bottom, we want (s+2) so thats why we use the first shift theorem and times the 4t by e^(-2t). now, can you do the same for the 1/(s+1)^3
P: 1,031
Okay so:

 4t will give you 4/s^2,but we don't want s at the bottom, we want (s+2) so thats why we use the first shift theorem and times the 4t by e^(-2t).
so would this mean:

$$t^2 would give: \frac{2}{s^3}$$

thus would:

$$\frac{2}{(s + 1)^3} be \frac{1}{2}t^2 e^{-t}$$

???

TFM
 P: 774 very good!!
 P: 1,031 Great So now: $$Y = \frac{1}{(s + 2)^3} + \frac{4}{(s + 2)^2}$$ this goes to: $$Y = \frac{1}{2}t^2 e^{-t} + 4te^{-2t}$$ So now we put this value of Y back into: $$s^2(Y) - 4 + 4(sY) + 4Y = \frac{1}{s + 2}$$ ??? TFM
 P: 774 y(t)=(1/2)t^2(e^-t) + 4t(e^-2t) This is the answer. dont substitute anything into anything.
 P: 1,031 Excellent. So $$y(t) = \frac{1}{2}t^2 e^{-t} + 4te^{-2t}$$ Tomorrow, I'll start (c), if thats okay? TFM
 P: 774 yep thats fine
 P: 1,031 Okay, so: (c): $$y'' + y = sin(t), y_0 = 1, y_0' = 0$$ L(y(t)) = Y L(y'(t)) = sL(y(t)) - y(0) L(y''(t)) = sL(y'(t))-y'(0) = s^2(L(y(t))-sy(0)-y'(0) there is no y', $$L(y(t)) = Y$$ $$L(y'(t)) = sL(y(t)) - 1$$ $$L(y''(t)) = sL(y'(t))-0 = s^2(L(y(t))-sy(0)-y'(0)$$ and for the sin(t) $$sin(\alpha t) = \frac{\alpha}{p^2 + \alpha^2}$$ since alpha = 1: $$sin(t) = \frac{1}{p^2 + 1}$$ Is this okay, I am not quite sure because the lack of y' TFM
 P: 774 That is absolutely right. what next?
 P: 1,031 Okay so: $$L(y(t)) = Y$$ $$L(y'(t)) = sL(y(t)) - 1$$ $$L(y''(t)) = s^2(L(y(t))-sy(0)$$ So now I substitute these into the original equation: $$y'' + y = sin(t)$$ $$s^2(L(y(t))-sy(0) + Y = \frac{1}{p^2 + 1}$$ Is this okay? Do I substitute y(0) into sy(0), or is there another value/leave it alone? TFM
 P: 774 That's fine substitute y(0) and also substitue the value L(y(t))
 P: 1,031 Okay, so $$s^2(L(y(t))-sy(0) + Y = \frac{1}{p^2 + 1}$$ $$s^2(Y)-1 + Y = \frac{1}{p^2 + 1}$$ $$Ys^2 + Y - 1 = \frac{1}{p^2 + 1}$$ Now I need to make the Y the subject: $$Ys^2 + Y = \frac{1}{p^2 + 1} + 1$$ factorise out $$Y(s^2 + 1) = \frac{1}{p^2 + 1} + 1$$ divide through: $$Y = \frac{\frac{1}{p^2 + 1} + 1}{s^2 + 1}$$ Now I need to find the inverse. Firstly, split it up into two fractions: $$Y = \frac{\frac{1}{p^2 + 1}}{s^2 + 1} + \frac{1}{s^2 + 1}$$ Does this look okay? TFM
 P: 774 in ur second line, you have: s^y-1+y=1/(s^2+1) the -1 on the left hand side should be -s (since you have:L(y'')=s(sy-1) and remember: s is p so dont put both...choose one of them :)
 P: 1,031 Okay so: $$s^2(Y)-1 + Y = \frac{1}{p^2 + 1}$$ $$Ys^2 + sY - 1 = \frac{1}{s^2 + 1}$$ Does this look okay now? TFM
 P: 774 you should have: Ys^2 - s + Y=1/(s^2+1) i think you made a mistake while substituting.
 P: 1,031 Okay, back a bit originally: $$s^2(L(y(t))-sy(0) + Y = \frac{1}{s^2 + 1}$$ y(0) = 1 L(y(t)) = Y put these in: $$s^2(Y) - 1 + Y = \frac{1}{s^2 + 1}$$ goes to: $$Ys^2 - 1 + Y = \frac{1}{s^2 + 1}$$ I still have a minus one where there should be a minus s.???? TFM

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